# Triangulations of submanifolds of smooth manifolds

Every smooth manifold $M$ has a PL structure, and therefore a triangulation. Given a submanifold $N$ of $M$, does anyone know some nice conditions for $N$ to be the subcomplex of some triangulation of $M$, or isotopic to one?

• Perhaps I am missing an important adjective, but I think your first sentence is false: arxiv.org/abs/1303.2354 – Adam Saltz May 10 '15 at 20:57
• @Adam--you missed the important adjective "smooth"; with that adjective it's a famous theorem of Whitehead. Manolescu's theorem says that there is a (high-dimensional) topological manifold without a simplicial triangulation. It was already known by Kirby-Siebenmann that there are topological manifolds that are not PL (more restrictive than just triangulable) and by Casson-Taubes-Freedman that there are non-triangulable topological 4-manifolds. – Danny Ruberman May 11 '15 at 0:27

It follows from Verona's solution to Thom's triangulation conjecture that the inclusion $N\hookrightarrow M$ is triangulable whenever it is proper and topologically stable, and $M$ and $N$ are without boundary.
• @IgorRivin: That's a good question, I confess to not being an expert. Here's what I know: $\operatorname{Emb}(M,N)$ is open in $C^\infty(M,N)$, and dense if $2m\le n$. Also, the space of topologically stable maps $M\to N$ is open and dense in $C^\infty(M,N)$ (the Thom-Mather theorem). I believe it follows that every embedding is isotopic to a topologically stable, hence triangulable, embedding. – Mark Grant May 12 '15 at 1:13
• Theorem 7.8 says "Let A be an a.s [abstract stratification] of finite depth. Then there exists a smooth triangulation $(K,\phi)$ of $A$ [then there's a bit more about how you can choose it to so that a certain map to a manifold is simplicial if you also want that]." Of course it takes some unwinding of the definitions to know what a smooth triangulation is and what an abstract stratification is. But it's basically a Thom-Mather space, and if $N$ is a proper smooth submanifold of $M$, I think the pair should satisfy the Thom-Mather conditions. – Greg Friedman Oct 31 '15 at 0:41